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(A) The domain of the function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is determined by $\frac{1+x}{1-x} > 0$,which implie

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$-1$

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(A) The domain of the function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is determined by $\frac{1+x}{1-x} > 0$,which implies $x \in (-1, 1)$.First,we simplify the expression for $f(x)$:$f(x) = \log(1+x) - \log(1-x) - 2x - \frac{x^3}{1-x^2}$.Now,we find the derivative $f'(x)$:$f'(x) = \frac{1}{1+x} + \frac{1}{1-x} - 2 - \frac{d}{dx} \left( \frac{x^3}{1-x^2} \right)$.$f'(x) = \frac{(1-x) + (1+x)}{1-x^2} - 2 - \frac{3x^2(1-x^2) - x^3(-2x)}{(1-x^2)^2}$.$f'(x) = \frac{2}{1-x^2} - 2 - \frac{3x^2 - 3x^4 + 2x^4}{(1-x^2)^2}$.$f'(x) = \frac{2(1-x^2) - 2(1-x^2)^2 - (3x^2 - x^4)}{(1-x^2)^2}$.$f'(x) = \frac{2 - 2x^2 - 2(1 - 2x^2 + x^4) - 3x^2 + x^4}{(1-x^2)^2}$.$f'(x) = \frac{2 - 2x^2 - 2 + 4x^2 - 2x^4 - 3x^2 + x^4}{(1-x^2)^2} = \frac{-x^4 - x^2}{(1-x^2)^2} = -\frac{x^2(x^2+1)}{(1-x^2)^2}$.Since $x^2(x^2+1) > 0$ for all $x \in (-1, 1) \setminus \{0\}$ and $(1-x^2)^2 > 0$,$f'(x) Thus,the function is decreasing on the entire interval $(-1, 1)$.Here,$a = -1$ and $b = 1$,so $\frac{a}{b} = \frac{-1}{1} = -1$.

If the interval in which the real-valued function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is decreasing is $(a, b)$,where $|b-a|$ is maximum,then $\frac{a}{b} =$

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